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An object undergoes simple harmonic motion in two mutually perpendicular directions, its position given by \(\vec{r}=A \sin \omega t \hat{\imath}+\) \(A \cos \omega t \hat{\jmath} .\) (a) Show that the object remains a fixed distance from the origin (i.e., that its path is circular), and find that distance. (b) Find an expression for the object's velocity. (c) Show that the speed remains constant, and find its value. (d) Find the angular speed of the object in its circular path.

Short Answer

Expert verified
The object remains a fixed distance \(A\) from the origin, which is also the radius of the circular path. The object's velocity is \(\vec{v}= A \omega \cos \omega t \hat{\imath} - A \omega \sin \omega t \hat{\jmath}\). The object's speed is constant and equal to \(A \omega\). The angular speed of the object in its circular path is \(\omega\).

Step by step solution

01

Determine the fixed distance from the origin

The fixed distance from the origin, or the radius of the circular path, can be obtained by finding the magnitude of the position vector \(\vec{r}\). This can be done by squaring and adding the two components and then taking the square root, similar to finding the hypotenuse of a right-angled triangle. For \(\vec{r}=A \sin \omega t \hat{\imath}+ A \cos \omega t \hat{\jmath}\), we get\[ \vec{r} \cdot \vec{r} = (A \sin \omega t \hat{\imath}+ A \cos \omega t \hat{\jmath} ) \cdot (A \sin \omega t \hat{\imath}+ A \cos \omega t \hat{\jmath} ) = A^2 \sin^2 \omega t + A^2 \cos^2 \omega t \]Using the identity \(\sin^2 x + \cos^2 x = 1\), we get\[ \vec{r} \cdot \vec{r} = A^2 \]So, \(| \vec{r} | = A \)
02

Find an expression for the object's velocity

The velocity is the derivative of position with respect to time. So, for each of the i and j components we have\[ \vec{v}= \frac{d \vec{r}}{dt} = A \omega \cos \omega t \hat{\imath} - A \omega \sin \omega t \hat{\jmath} \]
03

Show that the speed remains constant

The speed is the magnitude of the velocity vector, which we can find similarly to how we found the magnitude of the position vector:\[ \vec{v} \cdot \vec{v} = (A \omega \cos \omega t \hat{\imath} - A \omega \sin \omega t \hat{\jmath}) \cdot (A \omega \cos \omega t \hat{\imath} - A \omega \sin \omega t \hat{\jmath}) = A^2 \omega^2 \cos^2 \omega t + A^2 \omega^2 \sin^2 \omega t \]Using the identity \(\sin^2 x + \cos^2 x = 1\), we get\[ \vec{v} \cdot \vec{v} = A^2 \omega^2 \]So, \(| \vec{v} | = A \omega \)
04

Find the angular speed of the object

The angular speed is the speed divided by the radius of the circular path. Since the speed is \(A \omega\) and the radius is \(A\), we have\[ \omega_{angular} = \frac{\omega A}{A} = \omega \]

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Circular Motion
In simple harmonic motion, an object can exhibit circular motion due to its position vector. When dealing with the position vector \(\vec{r} = A \sin \omega t \hat{\imath} + A \cos \omega t \hat{\jmath}\), we see that it is composed of sine and cosine functions. These trigonometric functions represent the x and y components of motion in a plane.

When you compute the magnitude of this position vector, it's akin to finding the hypotenuse of a right triangle with sides \(A \sin \omega t\) and \(A \cos \omega t\). This results in:
  • \(\vec{r} \cdot \vec{r} = A^2 \sin^2 \omega t + A^2 \cos^2 \omega t\)
By applying the Pythagorean identity \(\sin^2 x + \cos^2 x = 1\), the magnitude simplifies to \(|\vec{r}| = A\). This shows that the object traces a circle with radius \(A\), keeping the distance from the origin constant as it moves.

This fixed distance is what we recognize as uniform circular motion—a key feature of simple harmonic motion.
Velocity and Speed
The velocity of an object in motion is a vector quantity that combines its speed and direction of travel. For this harmonic motion scenario, the velocity\(\vec{v}\) is found by differentiating the position vector with respect to time.

By taking the derivative, the velocity vector becomes:
  • \(\vec{v} = \frac{d \vec{r}}{dt} = A \omega \cos \omega t \hat{\imath} - A \omega \sin \omega t \hat{\jmath} \)
This indicates that the velocity acts tangentially to the circular path and changes direction as the object moves around the circle.

The speed, however, is the magnitude of this velocity vector, and it can be calculated as:
  • \(\vec{v} \cdot \vec{v} = A^2 \omega^2 \cos^2 \omega t + A^2 \omega^2 \sin^2 \omega t\)
Using the identity \(\sin^2 x + \cos^2 x = 1\), simplifies it to \(|\vec{v}| = A\omega\). This highlights that the speed of the object remains constant, irrespective of the specific position on its circular path.
Angular Speed
Angular speed refers to how fast the angle is changing as an object moves along a circular path. In this context, the angular speed is a constant value due to the simplicity of the harmonic motion.
  • The previously calculated speed for the object was \(A\omega\) and its circular path radius was \(A\).
The angular speed \(\omega_{angular}\) is determined by dividing the linear speed by the radius:
  • \(\omega_{angular} = \frac{A\omega}{A}= \omega\)
This expression reveals that the angular speed in harmonic motion remains equivalent to the original angular frequency of the motion \(\omega\).

This constancy signifies uniform angular velocity, showcasing that the object consistently rotates through its circle at the same rate, reflecting another hallmark of simple harmonic motion.
Mathematical Derivation
Mathematical derivation in the context of simple harmonic motion involves using trigonometric identities and derivatives to arrive at equations that explain physical phenomena. Here, we've seen this process in several key aspects.

Initially, the use of the Pythagorean identity \(\sin^2 x + \cos^2 x = 1\) was crucial for proving the circular path of the object's motion by confirming the fixed radius \(A\).

Next, by applying calculus through differentiation, we found the velocity expression from the position vector, demonstrating how calculus aids in transforming physical principles into usable formulas.
  • Differentiation revealed the tangential nature of velocity: \(\vec{v}= A \omega \cos \omega t \hat{\imath} - A \omega \sin \omega t \hat{\jmath}\).
  • Following this, derivation also facilitated finding constant speed \(A\omega\) and angular speed \(\omega_{angular} = \omega\).
These mathematical steps are the backbone of understanding these concepts, providing clarity and precision in modeling physical movements. Consistently, mathematical derivation underpins explanations and proofs that are pivotal in physics.

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Most popular questions from this chapter

The \(x\) - and \(y\) -components of motion of a body are both simple harmonic with the same frequency and amplitude. What shape is the path of the body if the component motions are (a) in phase, (b) \(\pi / 2\) out of phase, and (c) \(\pi / 4\) out of phase?

Two identical mass-spring systems consist of \(430-\mathrm{g}\) masses on springs of constant \(k=2.2 \mathrm{N} / \mathrm{m} .\) Both are displaced from equilibrium, and the first is released at time \(t=0 .\) How much later should the second be released so their oscillations differ in phase by \(\pi / 2 ?\)

A simple model for a variable star considers that the outer layer of the star is subject to two forces: the inward force of gravity and the outward force due to gas pressure. As a result, Newton's law for the star's outer layer reads \(m d^{2} r / d t^{2}=4 \pi r^{2} p-G M m / r^{2} .\) Here \(m\) is the mass of the outer layer, \(M\) is the total mass of the star, \(r\) is the star's radius, and \(p\) is the pressure. (a) Use this equation to show that the star's equilibrium pressure and radius are related by \(p_{0}=G M m / 4 \pi r_{0}^{4},\) where the subscript 0 represents equilibrium values. (b) As you'll learn in Chapter 18 , gas pressure and volume \(V\left(=\frac{4}{3} \pi r^{3}\right)\) are related by \(p V^{3 / 3}=p_{0} V_{0}^{5 / 3}\) (this is for an adiabatic process, a good approximation here, and the exponent \(5 / 3\) reflects the ionized gas that makes up the star). Let \(x=r-r_{0}\) be the displacement of the star's surface from equilibrium. Use the binomial approximation (Appendix A) to show that, when \(x\) is small compared with \(r,\) the righthand side of the above equation can be written \(-\left(G M m / r_{0}^{3}\right) x\) (c) since \(r\) and \(x\) differ only by a constant, the term \(d^{2}\) r/dt \(^{2}\) in the equation above can also be written \(d^{2} x / d t^{2} .\) Make this substitution, along with substituting the result of part (b) for the right- hand side, and compare your result with Equations 13.2 and 13.7 to find an expression for the oscillation period of the star. (d) What does your simple model predict for the period of the variable star Delta Cephei, with radius 44.5 times that of the Sun and mass of 4.5 Sun masses? (Your answer overestimates the actual period by a factor of about \(3,\) both because of oversimplified physics and because changes in the star's radius are too large for the assumption of a linear restoring force.)

A pendulum consists of a 320 -g solid ball \(15.0 \mathrm{cm}\) in diameter, suspended by an essentially mass-less string \(80.0 \mathrm{cm}\) long. Calculate the period of this pendulum, treating it first as a simple pendulum and then as a physical pendulum. What's the error in the simple-pendulum approximation? (Hint: Remember the parallel-axis theorem.)

The vibration frequency of a hydrogen chloride molecule is \(8.66 \times 10^{13} \mathrm{Hz}\). How long does it take the molecule to complete one oscillation?

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